What is the integral of (x^2+1/4 y^2+3) ?

The integral of (x^2+1/4 y^2+3) is 3x+(x^3)/3+1/4 y^2x+C

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integral of (x^2+1/4 y^2+3)

\int (x^{2} + \frac{1}{4} y^{2} + 3) d x

3 x + \frac{x ^{3}}{3} + \frac{1}{4} y^{2} x + C

Solution steps

\int x^{2} + \frac{1}{4} y^{2} + 3 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 d x + \int x^{2} d x + \int \frac{1}{4} y^{2} d x

\int 3 d x = 3 x

\int x^{2} d x = \frac{x ^{3}}{3}

\int \frac{1}{4} y^{2} d x = \frac{1}{4} y^{2} x

= 3 x + \frac{x ^{3}}{3} + \frac{1}{4} y^{2} x

Add a constant to the solution

= 3 x + \frac{x ^{3}}{3} + \frac{1}{4} y^{2} x + C

\frac{d y}{d t} = 0.03 y + 3600

\frac{\partial}{\partial x} (x^{\frac{3}{2}})

\int e^{b x} d x

3 cos (t)

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = - 1, y^{^{'}} (0) = - 2

What is the integral of (x^2+1/4 y^2+3) ?
The integral of (x^2+1/4 y^2+3) is 3x+(x^3)/3+1/4 y^2x+C